package algorithm.problems.tree;

/**
 * Created by gouthamvidyapradhan on 10/06/2017.
 * Accepted
 * You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.
 * <p>
 * The null node needs to be represented by empty parenthesis pair "()". And you need to omit all the empty parenthesis pairs that don't affect the one-to-one mapping relationship between the string and the original binary tree.
 * <p>
 * Example 1:
 * Input: Binary tree: [1,2,3,4]
 * 1
 * /   \
 * 2     3
 * /
 * 4
 * <p>
 * Output: "1(2(4))(3)"
 * <p>
 * Explanation: Originallay it needs to be "1(2(4)())(3()())",
 * but you need to omit all the unnecessary empty parenthesis pairs.
 * And it will be "1(2(4))(3)".
 * Example 2:
 * Input: Binary tree: [1,2,3,null,4]
 * 1
 * /   \
 * 2     3
 * \
 * 4
 * <p>
 * Output: "1(2()(4))(3)"
 * <p>
 * Explanation: Almost the same as the first example,
 * except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
 */
public class ConstructStringFromBinaryTree {
    public static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }

    /**
     * Main method
     *
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception {
        TreeNode t = new TreeNode(1);
        t.left = new TreeNode(2);
        t.left.left = new TreeNode(4);
        t.right = new TreeNode(3);
        System.out.println(new ConstructStringFromBinaryTree().tree2str(t));
    }

    public String tree2str(TreeNode t) {
        if (t == null) return "";
        String left = tree2str(t.left);
        String right = tree2str(t.right);
        if (left.equals("") && right.equals(""))
            return String.valueOf(t.val);
        if (left.equals(""))
            left = "()";
        else left = "(" + left + ")";
        if (!right.equals(""))
            right = "(" + right + ")";
        return t.val + left + right;
    }
}
